python方法具有1个变量的嵌套字典

Question

我发现很难命名这个问题，但它一定是一个简单的问题，而且我错过了一些基本的东西。

假设我有以下 Python dict() :

import json

dct = dict()
dct['hits'] = dict()
dct['hits']['hits'] = dict()
dct['hits']['hits']['a'] = 'b'
dct['hits']['hits']['b'] = 'c'
dct['aggregations'] = dict()
dct['aggregations']['a'] = 1
dct['aggregations']['b'] = 2

print(json.dumps(dct, indent=2))

{
  "hits": {
    "hits": {
      "a": "b",
      "b": "c"
    }
  },
  "aggregations": {
    "a": 1,
    "b": 2
  }
}

这可能看起来很熟悉，因为它是 ElasticSearch 返回结果的结构。

我正在构建一个使用该结果的函数。但有时我想访问dct['hits']['hits']，有时我想访问dct['aggregations']。

当然，我会使用一个带有变量的函数来建议我想要访问哪个字段，如下所示:

def foo(field):
    return dct[field]

如果 field='aggregations' 一切都很好。但是，当我希望字段为 ['hits']['hits'] 时，我该怎么办？

<小时/>

解决这个问题的一种方法(但它很丑陋)，迭代方法:

def foo(fields=('hits','hits')):
    wanted = dct
    for field in fields:
        wanted = wanted[field]
    return wanted

a = foo()
a
Out[47]: {'a': 'b', 'b': 'c'}
a = foo(('aggregations',))
a
Out[51]: {'a': 1, 'b': 2}

<小时/>

我尝试修改的实际功能:

def execute_scroll_query(es_client, query, indexes):
    try:
        response = es_client.search(index=indexes, scroll='2m', size=1000, body=query)
        scroll_size = len(response['hits']['hits'])
        sid = response['_scroll_id']
        while scroll_size > 0:
            try:
                for hit in response['hits']['hits']:
                    yield hit
                response = es_client.scroll(scroll_id=sid, scroll='2m')
                sid = response['_scroll_id']
                scroll_size = len(response['hits']['hits'])
            except Exception:
                print("Unexpected Exception while scrolling")
    except Exception:
        print("Unexpected Exception while fetching")

Answer 1

您可以使用functools.reduce，但在幕后，它使用迭代，并且可能不如显式迭代高效:

from functools import reduce

def foo(d, keys):
    return reduce(lambda x, y: x[y], keys, d)

foo(dct, ['hits', 'hits', 'a'])
#'b'